Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.6 Other Types of Equations and Applications - 1.6 Exercises - Page 146: 26


The solution is $x=-\dfrac{1}{4}$

Work Step by Step

$\dfrac{4x+3}{x+1}+\dfrac{2}{x}=\dfrac{1}{x^{2}+x}$ Take out common factor $x$ from the denominator of the fraction on the right side of the equation: $\dfrac{4x+3}{x+1}+\dfrac{2}{x}=\dfrac{1}{x(x+1)}$ Multiply the whole equation by $x(x+1)$: $x(x+1)\Big[\dfrac{4x+3}{x+1}+\dfrac{2}{x}=\dfrac{1}{x(x+1)}\Big]$ $(x)(4x+3)+2(x+1)=1$ Evaluate the indicated operations: $4x^{2}+3x+2x+2=1$ Take $1$ to the left side of the equation and simplify: $4x^{2}+3x+2x+2-1=0$ $4x^{2}+5x+1=0$ Solve by factoring: $(x+1)(4x+1)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $x+1=0$ $x=-1$ $4x+1=0$ $4x=-1$ $x=-\dfrac{1}{4}$ Since the original equation is undefined for $x=-1$, the solution is just $x=-\dfrac{1}{4}$
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